# Where is "countability" used in this proposition about product $\sigma$-algebra?

The following is a proposition about product sigma algebra from Folland's Real Analysis:

Proposition. If $$A$$ is countable, then $$\otimes_{\alpha\in A}M_{\alpha}$$ is is the $$\sigma$$-algebra generated by $$\{\Pi_{\alpha\in A}E_\alpha:E_\alpha\in M_\alpha\}$$.

Proof. If $$E_{\alpha}\in M_{\alpha}$$, then $$\pi_\alpha^{-1}(E_\alpha)=\Pi_{\beta\in A}E_\beta$$ where $$E_\beta=X_\beta$$ for $$\beta\not=\alpha$$; on the other hand, $$\Pi_{\alpha\in A}E_\alpha=\cap_{\alpha\in A}\pi^{-1}_\alpha(E_\alpha)$$. The result therefore follows from Lemma 1.1.

[Added]Lemma 1.1: If $$\mathcal{E}\subset M(\mathcal{F})$$ then $$M(\mathcal{E})\subset M(\mathcal{F})$$ where $$M(\mathcal{E})$$ denotes the sigma algebra generated by $$\mathcal{E}$$.

Where is the assumption "$$A$$ is countable" used in the proof?

• What does Lemma 1.1 say? Presumably the point is that the intersection of sets in a $\sigma$-algebra is again in the $\sigma$-algebra, as long as the intersection is over only countably many sets. Aug 12, 2014 at 19:19
• @AndresCaicedo: Thanks for your comment. I should have quoted Lemma 1.1, which would obviously not relate to the use of the "countability" assumption.
– user9464
Aug 12, 2014 at 20:56

On the last line, we can only guarantee that the element $\cap_{\alpha \in A} \pi^{-1}(E_\alpha)$ is an element of $\otimes_{\alpha}M_\alpha$ if $A$ is countable: the $\sigma$-algebra is closed under countable intersections, not necessarily arbitrary ones.